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Solve given equation log 2 3 plus log 2 a log 2 20?

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Updated: 12/23/2022

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Q: Solve given equation log 2 3 plus log 2 a log 2 20?

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Integral of cos2x log cosx-sinx coax plus since?

it is not possible to get the Integral of cos2x log cosx-sinx coax plus since there are no symbols given in the equation.

Solve Log f plus log 0.1 equals 6?

log(f) + log(0.1) = 6 So log(f*0.1) = 6 so f*0.1 = 106 so f = 107

What is the formula for finding the base and power of any integer?

It is my first answer. Is the problem to solve A=B^X ? where A and B are positive integers and X the power exponent of B The given equation can be rewritten in a logarithm form. Log A = X * Log B solving for a unique X X = Log A / Log B The result: Any positive integer A can be rewritten as a positive integer B to the distinct power X. Where X is Log A divided by Log B A = B ^(Log A / Log B) I think, this is the solution. Roger Verbeeck

Solve for x. 2log2x plus log21 equals log24?

I will assume that the "2" after each log is a subscript (indicating log to the base 2). Basically, you must use the well-known logarithmic identities, a log b = log ba, and log a + log b = log ab. 2 log2x + log21 = log24 log2x2 + log21 = log24 log2x2(1) = log24 log2x2 = log24 Take antilogs on both sides: x2 = 4 In this last equation, x is either 2 or -2. However, negative numbers are not appropriate for the original equation (assuming real numbers), so the only solution is 2. For safety, this should be checked in the original equation; I'll leave that part to you.

How do you solve logx minus 3 plus log 4 equals log x?

Use the identity log(ab) = log a + log b to combine the logarithms on the left side into a single term. Then take antilogarithms (just take the log away) on both sides.

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Related Questions

Integral of cos2x log cosx-sinx coax plus since?

it is not possible to get the Integral of cos2x log cosx-sinx coax plus since there are no symbols given in the equation.

What exponential equation is equivalent to the logarithmic equation e exponent a equals 47.38?

The given equation is exponential, not logarithmic!The logarithmic equation equivalent to ea= 47.38 isa = ln(47.38)ora = log(47.38)/log(e)The given equation is exponential, not logarithmic!The logarithmic equation equivalent to ea= 47.38 isa = ln(47.38)ora = log(47.38)/log(e)The given equation is exponential, not logarithmic!The logarithmic equation equivalent to ea= 47.38 isa = ln(47.38)ora = log(47.38)/log(e)The given equation is exponential, not logarithmic!The logarithmic equation equivalent to ea= 47.38 isa = ln(47.38)ora = log(47.38)/log(e)

What logarithmic equation is equivalent to 9x equals 27?

9x = 27 log(9) + log(x) = log(27) log(x) = log(27) - log(9) log(x) = log(27/9) 10log(x) = 10log(27/9) x = 27/9 x = 3 This strikes us as the method by which the federal government might solve the given equation ... after appointing commissions to study the environmental impact and recommend a method of solution, of course.

How do you solve a log?

You calculate a log, you do not solve a log!

How do you solve log1 plus log2 plus log3?

log1 + log2 + log3 = log(1*2*3) = log6

Solve Log f plus log 0.1 equals 6?

log(f) + log(0.1) = 6 So log(f*0.1) = 6 so f*0.1 = 106 so f = 107

How can you calculate pka valve if pH is given?

You can calculate the pKa value by using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where [A-] is the concentration of the conjugate base and [HA] is the concentration of the acid. Rearranging the equation, you can solve for pKa by taking the antilog of both sides after isolating pKa.

What is the formula for finding the base and power of any integer?

It is my first answer. Is the problem to solve A=B^X ? where A and B are positive integers and X the power exponent of B The given equation can be rewritten in a logarithm form. Log A = X * Log B solving for a unique X X = Log A / Log B The result: Any positive integer A can be rewritten as a positive integer B to the distinct power X. Where X is Log A divided by Log B A = B ^(Log A / Log B) I think, this is the solution. Roger Verbeeck

Solve for x. 2log2x plus log21 equals log24?

I will assume that the "2" after each log is a subscript (indicating log to the base 2). Basically, you must use the well-known logarithmic identities, a log b = log ba, and log a + log b = log ab. 2 log2x + log21 = log24 log2x2 + log21 = log24 log2x2(1) = log24 log2x2 = log24 Take antilogs on both sides: x2 = 4 In this last equation, x is either 2 or -2. However, negative numbers are not appropriate for the original equation (assuming real numbers), so the only solution is 2. For safety, this should be checked in the original equation; I'll leave that part to you.

How do you solve logx minus 3 plus log 4 equals log x?

Use the identity log(ab) = log a + log b to combine the logarithms on the left side into a single term. Then take antilogarithms (just take the log away) on both sides.

What is the log prosedure?

I assume you are asking how to solve a logarithmic equation. Well let's quickly review what the log function is: for the equation log(x)=y, we are saying that 10^y=x. So once you have isolated the logarithm, take the value of the base, raise it to the nth power (when 'n' is the value that the function is equal to) and set that equal to the value inside of the log.

How do you solve 2logx plus 3logx equals 10?

2 log(x) + 3 log(x) = 105 log(x) = 10log(x) = 10/5 = 210log(x) = (10)2x = 100

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